Question: Is ${869591}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {869591}= &&{8}\cdot100000+ \\&&{6}\cdot10000+ \\&&{9}\cdot1000+ \\&&{5}\cdot100+ \\&&{9}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {869591}= &&{8}(99999+1)+ \\&&{6}(9999+1)+ \\&&{9}(999+1)+ \\&&{5}(99+1)+ \\&&{9}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {869591}= &&\gray{8\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {8}+{6}+{9}+{5}+{9}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${869591}$ is divisible by $3$ if ${ 8}+{6}+{9}+{5}+{9}+{1}$ is divisible by $3$ Add the digits of ${869591}$ $ {8}+{6}+{9}+{5}+{9}+{1} = {38} $ If ${38}$ is divisible by $3$ , then ${869591}$ must also be divisible by $3$ Add the digits of ${38}$ $ {3}+{8} = \color{#9D38BD}{11} $ If $\color{#9D38BD}{11}$ is divisible by $3$ , then ${38}$ must also be divisible by $3$ $\color{#9D38BD}{11}$ is not divisible by $3$, therefore ${869591}$ must not be divisible by $3$.